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Weight in an Elevator and Normal Force

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A person who normally weighs 200 lbs is standing on a scale inside of an elevator. The elevator is moving downwards at a constant speed of 7.0 m/s and then begins to slow down at a rate of 4.9 m/s^2. Before the elevator begins to decelerate, the reading of the scale is ____, and while the elevator is slowing down, the scale reads:

    A. 200 lbs, 100 lbs.
    B. less than 200 lbs, 300 lbs
    C. less than 200 lbs, 100 lbs,
    D. 200 lbs, 300 lbs.
    E. 200 lbs, 200 lbs

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    All a scale knows is the normal force being exerted on it.
    Initially there's no acceleration so weight is the same. That eliminates B and C. It then begins to decelerate at 4.9 m/s^2. I thought either which way acceleration is negative. Your velocity is going from being positive to less positive.

    F - mg = ma
    F = mg + ma
    F = 200 lbs (9.8 m/s^2 - 4.9 m/s^2)
    I know the units aren't right and my teacher said I wouldn't have to convert but I don't know why I would ADD 4.9 m/s^2. I thought the problem was just giving the magnitude of the deceleration.
     
  2. jcsd
  3. Mar 1, 2012 #2
    if it's slowing down, in which direction does that imply the acceleration is pointed?

    also in questions like this it is possible to use your experience to answer it, that is, if you've got enough experience with elevators.
     
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3
    Ok. I see. If velocity is moving down, in order for it to be slowing down, acceleration has to point in the opposite direction, up.
     
  5. Mar 1, 2012 #4
    yep ;)

    always make sure you pay close attention to the wording of a problem in order to determine the directions in which the vectors are pointed
     
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